What is the matrix used to find the reflected (x, y) coordinate in the line y=mx?

Question

I hope this makes sense, I'm essentially looking for a matrix in which you can just substitute in the gradient m from y=mx and find the reflected coordinates? If this doesn't make any sense please say why? Regards Tom

Answer 1

The matrix you are searching is:

$$ \dfrac{1}{1+m^2} \left[ \begin{array}{cccc} 1-m^2&2m\\ 2m & m^2-1 \end{array} \right] $$ For a proof see may answer to Linear Transformation with eigenvectors and eigenvalues

Answer 2

Your question does make sense, since you describe a linear transformation. Any reflection about a line through the origin is a linear transformation.

One way to find the transformation is to find the image of the point $(1,0)$ [call it $(a,b)$] and the image of the point $(0,1)$ [call it $(c,d)$]. The matrix is then

$$\begin{bmatrix} a & c \\ b & d \\ \end{bmatrix}$$

Here is one way to find $(a,b)$. The line between $(1,0)$ and $(a,b)$ is perpendicular to the line $y=mx$: what does that say about the slope of the line between $(1,0)$ and $(a,b)$? Also, the midpoint of the line segment between $(1,0)$ and $(a,b)$ is on the line $y=mx$.

Each of those two statements gives you a linear equation in $a$ and $b$. Solve those two simultaneous equations and you have your $a$ and $b$. Do almost the same thing to find $c$ and $d$.

There are other ways to find your matrix, of course, but this way uses no advanced concepts.